1 g of water, of volume 1 cm3at 100°C, is converted into steam at same temperature under normal atmospheric pressure (1×105 Pa). The volume of steam formed equals 1671cm3. If the specific latent heat of vaporization of water is 2256 J/g, the change in internal energy is:

ΔQ=2256×1=2256 J ΔW=PVsteam -Vwater =1×105[1671-1]×10-6=1670×105×10-6=167 JBy first law of thermodynamics: As ΔQ=ΔU+ΔW2256=ΔU+167ΔU=2089 J

1 g of water, of volume 1 cm3at 100°C, is converted into steam at same temperature under normal atmospheric pressure (1×105 Pa). The volume of steam formed equals 1671cm3. If the specific latent heat of vaporization of water is 2256 J/g, the change in internal energy is:

ΔQ=2256×1=2256 J ΔW=PVsteam -Vwater =1×105[1671-1]×10-6=1670×105×10-6=167 JBy first law of thermodynamics: As ΔQ=ΔU+ΔW2256=ΔU+167ΔU=2089 J

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1 g of water, of volume $1 {cm}^{3}$ at 100°C, is converted into steam at same temperature under normal atmospheric pressure $(1 \times 10^{5} Pa) .$ The volume of steam formed equals $1671 {cm}^{3}$ . If the specific latent heat of vaporization of water is 2256 J/g, the change in internal energy is:

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2256 J

2423 J

2089 J

167 J

answer is C.

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Detailed Solution

$ΔQ = 2256 \times 1 = 2256 J ΔW = P [V_{steam} - V_{water}] \begin{array}{l} = 1 \times 10^{5} [1671 - 1] \times 10^{- 6} \\ = 1670 \times 10^{5} \times 10^{- 6} \\ = 167 J \end{array}$

By first law of thermodynamics:

$As ΔQ = ΔU + ΔW$

$2256 = ΔU + 167$

$ΔU = 2089 J$

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1 g of water, of volume 1 cm3at 100°C, is converted into steam at same temperature under normal atmospheric pressure (1×105 Pa). The volume of steam formed equals 1671cm3. If the specific latent heat of vaporization of water is 2256 J/g, the change in internal energy is:

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